There is a popular (and I think helpful) example of etale covers, namely covers of Riemann surfaces with ramification points removed. Is there a similarly accessible example to motivate Nisnevich covers?

Personally, I have found examples of Nisnevich covers un-enlightening. The most important property is: $p: U \to X$ is a Nisnevich cover if and only if it is an étale cover, and there exists a sequence $X \supset Z_1 \supset Z_2 \supset ... \supset Z_n = \varnothing$ of closed subschemes such that $p$ has a section over each $Z_i - Z_{i + 1}$. This allows induction arguments. And of course another motivation for Nisnevich is that every Nisnevich cover of a field has a section, and so the cohomological dimension of a "point" $Spec(k)$ is zero - a "failing" of the étale topology.
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nameJul 27 '12 at 1:50

1 Answer
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Well, a representative example is where you take some arbitrary etale cover Y of X which splits over a closed subvariety Z of X, then form the Nisnevich cover of X consisting of the open complement X - Z together with the open subscheme Y' of Y where you remove all but one of the copies of Z lying above Z. For instance Z could be a point, and our field could be algebraically closed.

The intuition I find helpful is that descent for the Nisnevich topology is meant to be an easier-to-precisely-phrase consequence of the principle "X is gotten by gluing X-Z to a tubular neighborhood of Z, along the punctured tubular neighborhood of Z". The idea being that, in the situation of the previous paragraph, the tubular neighborhoods of Z in X and Z in Y should be the same, Y --> X being etale.

As an example of this intuition, note that if X is a smooth variety of dimension n (over an algebraically closed field for simplicity) and x is a point of X, then by choosing n independent parameters at x you can find a Zariski neighborhood X' of x and an etale map X' --> A^n which, together with A^n - 0 ---> A^n, makes a Nisnevich cover of A^n with intersection equal to X'-x. This corresponds (ish) to the fact that the tubular neighborhood of x in X should be the same as the tubular neighborhood of 0 in A^n.